1 00:00:00,000 --> 00:00:00,000 2 00:00:00,000 --> 00:00:08,820 PROFESSOR: Welcome back to recitation. 3 00:00:08,820 --> 00:00:11,560 In this video, what I'd like us to do is answer the 4 00:00:11,560 --> 00:00:13,400 following question. 5 00:00:13,400 --> 00:00:14,950 Suppose that f is a continuous, 6 00:00:14,950 --> 00:00:16,730 differentiable function. 7 00:00:16,730 --> 00:00:20,730 And if it's derivative, if f prime is never 0, and a is not 8 00:00:20,730 --> 00:00:24,680 equal to b, then show that f of a is not equal to f of b. 9 00:00:24,680 --> 00:00:26,940 I'm going to let you think about it for a while, see if 10 00:00:26,940 --> 00:00:28,850 you can come up with a good reason for that, and then I'll 11 00:00:28,850 --> 00:00:30,140 be back to explain my reasons. 12 00:00:30,140 --> 00:00:38,160 13 00:00:38,160 --> 00:00:38,510 OK. 14 00:00:38,510 --> 00:00:42,890 Our object, again, is to show, if f is a continuous and 15 00:00:42,890 --> 00:00:46,550 differentiable and its derivative is never 0 and 16 00:00:46,550 --> 00:00:49,580 you're looking at two x values that are different, show that 17 00:00:49,580 --> 00:00:51,480 their y values have to be different. 18 00:00:51,480 --> 00:00:53,430 Show that if the inputs are different, the outputs have to 19 00:00:53,430 --> 00:00:54,300 be different. 20 00:00:54,300 --> 00:00:57,010 Now, this might remind you of something you saw in lecture 21 00:00:57,010 --> 00:01:00,720 about if the derivative has a sign, show the function, if 22 00:01:00,720 --> 00:01:02,420 the derivative is positive, show the function is always 23 00:01:02,420 --> 00:01:04,200 increasing, or if the derivative is negative, show 24 00:01:04,200 --> 00:01:06,160 the function is always decreasing. 25 00:01:06,160 --> 00:01:10,340 So this is a similar type of problem to that. 26 00:01:10,340 --> 00:01:12,470 So what we're going to use is actually 27 00:01:12,470 --> 00:01:13,880 the mean value theorem. 28 00:01:13,880 --> 00:01:15,690 If you'll notice, I have f. 29 00:01:15,690 --> 00:01:18,560 It does satisfy the mean value theorem on an 30 00:01:18,560 --> 00:01:20,220 interval from a to b. 31 00:01:20,220 --> 00:01:22,420 I haven't even specified which is bigger, a or b. 32 00:01:22,420 --> 00:01:25,350 But it doesn't matter in this case. 33 00:01:25,350 --> 00:01:26,280 So what do we know? 34 00:01:26,280 --> 00:01:31,290 The mean value theorem tells us that if we look at-- well, 35 00:01:31,290 --> 00:01:37,990 let's just write it out-- f of b minus f of a over b minus a 36 00:01:37,990 --> 00:01:40,610 is equal to f prime of c-- 37 00:01:40,610 --> 00:01:43,200 and what do we know-- for c between a and b. 38 00:01:43,200 --> 00:01:46,100 39 00:01:46,100 --> 00:01:49,030 So we want to know whether or not f of b minus f of 40 00:01:49,030 --> 00:01:50,320 a can ever be 0. 41 00:01:50,320 --> 00:01:52,740 We're trying to show that it cannot be 0. 42 00:01:52,740 --> 00:01:57,960 So we're going to isolate this expression and show that this 43 00:01:57,960 --> 00:01:59,800 subtraction cannot be 0. 44 00:01:59,800 --> 00:02:00,530 Well, how do we do that? 45 00:02:00,530 --> 00:02:03,670 Let me come over here to give us a little more room. 46 00:02:03,670 --> 00:02:05,260 I'm going to rewrite the mean value theorem. 47 00:02:05,260 --> 00:02:07,331 I'm going to multiply through by b minus a. 48 00:02:07,331 --> 00:02:12,250 49 00:02:12,250 --> 00:02:18,460 So we get f prime of c time b minus a. 50 00:02:18,460 --> 00:02:20,740 Now, we just want to show, again, that f of b minus f of 51 00:02:20,740 --> 00:02:22,530 a cannot be 0. 52 00:02:22,530 --> 00:02:25,070 What's the only thing-- well, not only thing, we know two 53 00:02:25,070 --> 00:02:26,570 things-- what two things do we know? 54 00:02:26,570 --> 00:02:29,350 We know f prime of c is not 0. 55 00:02:29,350 --> 00:02:30,380 That was given to you. 56 00:02:30,380 --> 00:02:33,360 f prime is never 0, so certainly at any fixed value, 57 00:02:33,360 --> 00:02:35,360 f prime of c is not 0. 58 00:02:35,360 --> 00:02:37,860 So we know this term is not 0. 59 00:02:37,860 --> 00:02:40,530 We also know that b is not equal to a, so we know b minus 60 00:02:40,530 --> 00:02:42,160 a is not 0. 61 00:02:42,160 --> 00:02:45,580 The only way to get a product of two numbers to be 0 is if 62 00:02:45,580 --> 00:02:46,380 one of them is 0. 63 00:02:46,380 --> 00:02:51,080 So this in fact, this product is not equal to 0. 64 00:02:51,080 --> 00:02:53,830 The fact that this product is not equal to 0 tells us f of b 65 00:02:53,830 --> 00:02:55,950 minus f of a is not equal to 0. 66 00:02:55,950 --> 00:02:59,970 And that alone is enough to conclude that f of b is not 67 00:02:59,970 --> 00:03:03,150 equal to f of a. 68 00:03:03,150 --> 00:03:06,200 So, again, let me just point out of this is probably 69 00:03:06,200 --> 00:03:08,580 reminds you very much of the type of thing you've seen 70 00:03:08,580 --> 00:03:12,140 where you were showing if f prime had a sign, then you 71 00:03:12,140 --> 00:03:15,400 could determine whether f was increasing or decreasing. 72 00:03:15,400 --> 00:03:17,200 It's the same type of problem as that. 73 00:03:17,200 --> 00:03:20,180 It's exploiting what the mean value theorem tells you. 74 00:03:20,180 --> 00:03:22,080 So I think we'll stop there. 75 00:03:22,080 --> 00:03:22,335